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I would like to know how I can find the result of $1761^3 \bmod 7$. Is there any rule?

Thanks so much for your help!

hmakholm left over Monica
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Suslik
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    $a^n\mod{p}\equiv(a\mod{p})^n$ – kingW3 Aug 10 '14 at 12:43
  • @kingW3 $\ $ That's not valid notation. Probably you meant $\tag{}$ either $\ a^n\equiv, (a\ {\rm mod}\ p)^n\pmod p$ $\tag{}$ or $\ a^n\ {\rm mod}\ p, =, (a\ {\rm mod}\ p)^n, {\rm mod}\ p\ \ $ – Bill Dubuque Aug 10 '14 at 15:34

5 Answers5

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The remainder is the same as the remainder of $1761/7$ to the third.

You should be able to manage from there.

Eutherpy
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Let me change 1761 to 2014 to leave something to you.

Since $98=14\times7$, we have $100a=2a\mod7$ for any integer $a$. Thus $2014=14+2\times20=54=5\mod7$, and so $2014^3=5^3=125=25+2=27=-1\mod 7$.

Joonas Ilmavirta
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You should make use of the fact that if $a \equiv b \bmod c$, then $a^n \equiv b^n \bmod c$.

In this case, $a = 1761$, $c = 7$, and $n = 3$. What about $b$? There are infinitely many $b$ such that $1761 \equiv b \bmod 7$, but to make our calculation easier, we might as well take $b$ to be a small number. More precisely, there is a unique $b \in \{0, 1, \dots, 6\}$ such that $1761 \equiv b \bmod 7$; I suggest using this $b$.

Michael Albanese
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$\begin{eqnarray}{\bf Hint}\ \ {\rm mod}\ 7\!:\ \color{#0a0}{10}\equiv \color{#c00}3\,\Rightarrow\, 1761 =&& \color{#0a0}{10}^3+7\cdot\color{#0a0}{10}^2\!+6\cdot \color{#0a0}{10}+1\\ \equiv && \ \ \color{#c00}3^3 + 0\cdot\color{#c00}3^2\,-\,1\cdot \color{#c00}3\,+\,1\\ \equiv && \ \ 6\ \ -\ \ 0\ \ \ \ -\ \ \ \ 3\ \ +\ \ 1 \,\equiv\, 4 \end{eqnarray}$

Therefore $\,1761\equiv 4\,\overset{\rm\color{#c0f}{CPR}}\Rightarrow\, 1761^{\large 3}\equiv 4^{\large 3}\equiv 4(16)\equiv 4(2)\equiv 1$

where above we used $\rm\color{#C0F}{CPR} = $ Congruence Power Rule: $\ a\equiv b\,\Rightarrow\, a^{\large n}\equiv b^{\large n}\pmod m$

Community
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Bill Dubuque
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The key here is that if $a \equiv b (\mod n)$ and $a' \equiv b' (\mod n)$, then $a\cdot a' \equiv b \cdot b' (\mod n)$. A general strategy for finding $X (\mod n)$ is to find factors of $X$ which are equivalent to smaller numbers modulo $n$. Because of the multiplicative property stated above, you can replace the factors of $X$ with these smaller numbers without changing the remainder. Repeatedly apply this procedure, until you have shown that $X$ is equivalent to a number between $0$ and $n-1$. I'll illustrate this below.

Find the remainder when $2275^9$ is divided by $3$:

$2275=5^2\cdot 7\cdot 13.$ Now $5 \equiv 2 (\mod 3)$, $7 \equiv 1 (\mod 3)$, and $13 \equiv 1 (\mod 2)$. It follows that $2275 \equiv 2^2\cdot 1 \cdot 1 \equiv 1 (\mod 3)$. Therefore we see that $2275^9 \equiv 1^9 \equiv 1 (\mod 3)$, so the remainder when $2275$ is divided by $3$ is $1$.

You can sometimes also make use of Fermat's little theorem or other handy tricks to solve such problems, but I think the above describes a good general strategy.

vociferous_rutabaga
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